Solutions to Exercises on Rotational Mechanics HC Verma's Concepts of Physics Part 1 (Q58-Q62)

 Q#58

Suppose the platform with the kid in the previous problem is rotating in an anticlockwise direction at an angular speed of ⍵. The kid starts walking along the rim with a speed v relative to the platform also in the anticlockwise direction. Find the new angular speed of the platform.  

Answer:
The initial angular momentum of the system = (I + MR²)⍵
Let the new angular speed of the platform = ⍵'
The speed of the rim = ⍵'R
Hence the speed of the boy with respect to the ground = ⍵'R+v
Angular momentum of the boy = M(⍵'R + v)R
The angular momentum of the platform = I⍵'
Since the total angular momentum will be conserved,

I⍵' + M(⍵'R + v)R = (I + MR²)⍵
I⍵' + M⍵'R² + MvR = (I + MR²)⍵
(I + MR²)⍵' = (I + MR²)⍵ – MvR
⍵' = {(I + MR²)⍵ – MvR}/(I + MR²)
⍵' = ⍵ – MvR/(I + MR²)

Q#59
A uniform rod of mass m and length l is struck at an end by a force F perpendicular to the rod for a short time interval t. Calculate (a) the speed of the center of mass, (b) the angular speed of the rod about the center the center of mass, (c) the kinetic energy of the rod and (d) the angular momentum of the rod about the center of mass after the force has stopped to act. Assume that t is so small that the rod does not appreciably change its direction while the force acts.

Answer:
(a) The impulse of the force = Ft
The initial speed of the center of mass = u =0
If the final speed of the center of mass is v, then
Change of linear momentum = Impulse

mv – m 0 = Ft
mv = Ft
v = Ft/m

(b) Let the angular speed of the rod about the center of mass =⍵
The moment of inertia of the rod about the CoM and perpendicular to the rod = I = ml²/12
Final angular momentum = I⍵
Initial angular momentum = zero
The torque of the force about the center of mass T = Fl/2
The angular impulse = Tt = Flt/2
The change in angular momentum = angular impulse

I⍵ – 0 = Flt/2
⍵ = Flt/2I = Flt/2(ml²/12)

⍵ = 6Ft/ml

(c) The kinetic energy of the rod
Linear kinetic energy + Rotational kinetic energy

K = ½mv² + ½I⍵²
K = ½m(Ft/m)² + ½(ml²/12)(6Ft/ml)²
= F²t²/2m + 36F²t²/24m
K = 2F²t²/m

(d) The angular momentum

L = I⍵ = (ml²/12)(6Ft/ml)
= 6Ftml²/12ml
L = Ftl/2

Q#60
A uniform rod of length L lies on a smooth horizontal table. A particle moving on the table strikes the rod perpendicularly at an end and stops. Find the distance traveled by the center of the rod by the time it turns through a right angle. Show that if the mass of the rod is four times that of the particle, the collision is elastic.  

Answer:
Let the mass of the particle and the rod be m and M respectively. If the velocity of the particle is v initially and the angular speed of the rod after the strike = ⍵.
Taking the rod and the particle as a system, there is no external force or torque on the system.

t = time taken in turning the rod through a right angle
The linear and angular momentum will be conserved. If the velocity of the center of mass after the strike is v', then

Mv' = mv
v' = mv/M                                       (i)

The angular momentum of the particle about the center = mvL/2
The angular momentum of the rod = I⍵ = ML²⍵/12
Equating the two,

ML²⍵/12 = mvL/2
ML⍵/6 = mv
⍵ = 6mv/ML

Time taken in turning through a right angle

t = π/2⍵ = πML/12mv

The distance traveled by the center of mass of the rod in this time

= v't
= (mv/M)(πML/12mv)
= πL/12

When M = 4m
v' = mv/4m = v/4
And, ⍵ = 6mv/ML = 6mv/4mL = 3v/2L
Initial kinetic energy of the system = ½mv²

Final kinetic energy of the system = Translational kinetic energy + Rotational kinetic energy

= ½Mv'² + ½I⍵²
= ½(4m)v²/16 + ½(ML²/12)(3v/2L)²
= mv²/8 + (4mL²/24)(9v²/4L²)
= mv²/8 + 9mv²/24
= 12mv²/24
= ½mv²

= Initial Kinetic energy of the system

Since the initial and final kinetic energy of the system is equal hence the collision is elastic.

Q#60
Suppose the particle of the previous problem has a mass m and a speed v before the collision and it sticks to the rod after the collision. The rod has a mass M. (a) Find the velocity of the center of mass C of the system constituting "the rod plus the particle". (b) Find the velocity of the particle with respect to C before the collision. (c) Find the velocity of the rod with respect to C before the collision. (d) Find the angular momentum of the particle and of the rod about the center of mass C before the collision. (e) Find the moment of inertia of the system about the vertical axis through the center of the mass C after the collision. (f) Find the velocity of the center of mass C and the angular velocity of the system about the center of mass after the collision.

ANSWER:
(a) Let the velocity of the CoM of the system be v'.
The linear momentum will be conserved. so,

(m + M)v' = mv
v' = mv/(m + M)

(b) The velocity of the particle w.r.t. CoM
v" = v – v'
= v – [mv/(m+M)]
= (mv + Mv – mv)/(m + M)
= Mv/(m+M)

(c) Let the velocity of the rod w.r.t C before collision = V

V = Velocity of the rod – velocity of CoM
= 0 – v'
= –mv/(m + M)

(d) Let the distance of the C from the particle before the collision = x

x = [m(0) + M(L/2)]/(m + M)
= ½ ML/(m+M)

The angular momentum of the particle about

L_c = mv"[½ ML/(m+M)]
L_c = {mMv/(m + M)}{½ML/(m + M}
L_c = ½ mM²vL/(m + M)²
(Anticlockwise)

The distance of C from the center of the rod,
y = (L/2) – x 
= (L/2) – ML/2(m + M)
= (mL + ML – ML)/2(m+M)
y = ½ mL/(m + M)

The angular momentum of the rod about
L_C = MV[½ mL/(m + M)]
L_C = {Mmv/(m + M)}{½ mL/(m + M)}
L_C = ½ Mm²vL/(m + M)²
(Anticlockwise)

(e) The moment of inertia of the system about C after the collision

I_C = M.I. of the particle about C + M.I. of the rod about C
= mx² + ML²/12 + My²
I_C = mM²L²/4(m + M)² + ML²/12 + Mm²L²/4(m + M)²
= {3mML²(M + m) + ML²(m + M)²}/12(m + M)²
= {ML²(M + m)(3m + m + M)}/12(m + M)²
I_C = ML²(4m + M)/12(m + M)

(f) The velocity of the CoM of the rod plus the particle has been calculated in (a) = mv/(M + m). It can also be calculated by finding the velocity of the CoM of the system just before the collision because there is no external force on the system so the velocity of the center of mass before the collision will be the same as after the collision.

The velocity of CoM before the collision
= (M x 0 + mv)/(M + m)
= mv/(M + m)

The angular momentum will also be conserved. The angular momentum after the collision = the angular momentum before the collision

= The angular momentum of the particle about C + The angular momentum of the rod about C

= mM²vL/2(m + M)²+ Mm²vL/2(m + M)²

[As calculated in (d)]

= ½mMvL(M + m)/(M + m)²
= ½mMvL/(M + m)
Let the angular speed of the rod be ⍵.
Then ⍵ = Angular Momentum/M.I

= {mMvL/2(M + m)}/{ML²(4m + M)/12(m + M)}

[M.I. of the system calculated in (e)]
= 6mv/{L(4m + M)}


Q#62
Two small balls A and B, each of mass m, are joined rigidly by a light horizontal rod of length L. The rod is clamped at the center in such a way that it can rotate freely about a vertical axis through its center. The system is rotated with an angular speed ⍵ about the axis. A particle P of mass m kept at rest sticks to the ball A as the ball collides with it. Find the new angular speed of the rod.  

Answer:
The moment of inertia of the ball system,

I = 2m(L/2)² = 2mL²/4 = mL²/2

The angular momentum of the system

L = I⍵ = mL²⍵/2

When the particle P of mass m sticks to the ball A the moment of inertia of the system changes.
The new moment of inertia

I' = I + m(L/2)² = mL²/2 + mL²/4 = 3mL²/4

Let the new angular speed be ⍵'
Since the angular momentum will be conserved, hence

I'⍵' =I⍵
(3mL²/4)⍵' = (mL²/2)⍵
3⍵'/4 = ⍵/2
⍵' = 2⍵/3 

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